Question

Find the rate of change of the distance between the origin and a moving point on the graph of y=x^2+1 if dx/dt= 2 centimeters per second

Answer 1

Well, you always need a formula that can relate all the information given. In this case, all you're really given is a function and "distance", so the best place to start is with setting up the distance formula. So always with these problems, we try and derive the formula first, so here is distance formula:

\[s = \sqrt{x^{2}+ y^{2}}\]because I have no coordinate points to work with, I left out tHE (x-x1)^2 part of it. So we have our formula, normally derivative would come 2nd. But we have to realize that if we took the derivative of this, we would have a dy/dx that cannot be solved for. This means we have to eliminate y somehow. In general, you usually want to reduce your formula down to as few variables as necessary. In this case we can replace y. We were originally told that y = x^2 + 1, so thats exactly what im going to replace y with

\[s = \sqrt{x^{2}+ (x^{2}+1)^{2}} = \sqrt{x^{2}+ x^{4}+ 2x^{2} + 1} = \sqrt{x^{4}+ 3x^{2} + 1}\]So finally we've reduced theformula down to only one variable and simplified. Now we can take our derivative.
\[\frac{ ds }{ dt }= \frac{ 1 }{ 2 }(x^{4}+ 3x^{2}+1)^{-1/2}(4x^{3}\frac{ dx }{ dt }+ 6x \frac{ dx }{ dt }) \]Everytime I took the derivative of an x-term, I added dx/dt to it. Youw ould do this with all variables. Take the derivative of a variable, tack on d?/dt basically. So now I can plug in dx/dt = 2 and simplify to get:
\[\frac{ ds }{ dt }= \frac{ 4x^{3}+6x }{ \sqrt{x^{4}+3x^{2}+1} }\]
And since there are no coordinates to actually plug in, this is the answer.

This problem is out of the larson textbook, yes?

Answer 2

That is the answer I got but in the text book the numerator is [2(2x^3+1)] I just don't know where that 2 cAme from

Answer 3

Odd. I mean, I probably have your same textbook because its in the one I have. You might have a different edition, though. But yeah, the back of my book says the answer we both have...to the exact same question. So why your book would say differently, im not sure.

Answer 4

I know what is wrong..... You have to substitute dx/dt=2!!!!

Answer 5

I did that actually, lol.

Answer 6

But then you get 8x^3+12x and then tKe out a 2